What does a lower bound do to the degrees of freedom of this random vector? Consider two procedures to sequentially draw $4$ random numbers $(k_1, k_2, k_3, k_4)$ that add up to $100$:


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*Procedure 1 can pick real numbers $\geq 0$

*Procedure 2 can pick real numbers $\geq 20$
Procedure 1 will draw $k_1\in(0,100)$, then $k_2\in(0,100-k_1)$, then $k_3\in(0,100-k_1-k_2)$, and finally $k_4\in(0,100-k_1-k_2-k_3)$.
Procedure 2 will draw $k_1\in(20,40)$, then $k2∈(20,60−k1)$, then $k3∈(20,80−k1−k2)$, and finally $k4∈(20,100−k1−k2−k3)$.
At each step, Procedure 1 needs to take into account previous draws and it describes a random vector with $4-1$ degrees of freedom since $k_4$ is determined by the first three draws. But Procedure 2 additionally has to worry about the number of draws that are still to come, due to the non-zero lower bound. 
How can I describe the difference between these procedures more formally, perhaps in terms of degrees of freedom?
 A: Both procedures have three degrees of freedom, because the result stays in a three dimensional space (this is quite the definition of degrees of freedom). In both cases the sample space is a simplex, but in the first case the simplex is contiguous to the origin and has edges of length 100, in the second case instead the simplex has edges of length 20, and the vertex is in the point (20, 20, 20). So, as the dimensions are still 3, the number of degrees of freedom doesn't change.
Note that it's not obvious what probability each possible sample value should have. For instance, if you draw the first three values sequentially in the way you described, using a uniform probability on the possible values for each $k_i$, then the expected values of the first values will be higher than the ones following. In symbols: $E[k_1] > E[k_2] > E[k_3] = E[k_4]$. This is not necessary and other sampling distributions can avoid this effect. A notorious one that applies to n-dimensional simplexes is Dirichlet distribution.
To stress the formal difference of these two sampling spaces that you invented you can say that the first one has greater volume than the second. The first volume is $5^3$ times the second.
Another much simpler way to compare the same two procedure is to note that they are linearly related. If we call $X$ the random vector $(k_1, k_2, k_3, k_4)$ under the first procedure, and $X'$ the same vector under the second procedure, then $X' = X/5+20$, as long as their distributions are comparable. In any case, this applies to their domain, as discussed above. 
